Given { F(x) = 2x^2 - 3x $}$, Determine { F'(x) $}$ Using The First Principles.Determine:1. { \frac{dy}{dx}$}$ If { Y = 4x^5 - 6x^4 + 3x $} 2. \[ 2. \[ 2. \[ D_x\left[-\frac{\sqrt[3]{x}}{2} +

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Introduction

Differentiation is a fundamental concept in calculus that deals with the study of rates of change and slopes of curves. It is a crucial tool in various fields, including physics, engineering, and economics. In this article, we will explore the concept of differentiation using first principles, which is a method of finding the derivative of a function. We will also apply this concept to determine the derivative of a given function and find the derivative of a more complex function.

What is Differentiation Using First Principles?

Differentiation using first principles is a method of finding the derivative of a function by using the definition of a derivative. The definition of a derivative is given by:

fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

This definition states that the derivative of a function f(x) at a point x is equal to the limit of the difference quotient as h approaches zero.

Step-by-Step Guide to Differentiation Using First Principles

To differentiate a function using first principles, we need to follow these steps:

  1. Write the definition of a derivative: Write the definition of a derivative, which is given by:

    fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

  2. Substitute the function: Substitute the given function into the definition of a derivative.

  3. Simplify the expression: Simplify the expression by combining like terms and canceling out any common factors.

  4. Evaluate the limit: Evaluate the limit as h approaches zero.

Example 1: Differentiating a Quadratic Function

Let's consider the quadratic function:

f(x)=2x2βˆ’3xf(x) = 2x^2 - 3x

We want to find the derivative of this function using first principles.

Step 1: Write the definition of a derivative

The definition of a derivative is given by:

fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

Step 2: Substitute the function

Substitute the given function into the definition of a derivative:

fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

fβ€²(x)=lim⁑hβ†’02(x+h)2βˆ’3(x+h)βˆ’(2x2βˆ’3x)hf'(x) = \lim_{h \to 0} \frac{2(x + h)^2 - 3(x + h) - (2x^2 - 3x)}{h}

Step 3: Simplify the expression

Simplify the expression by combining like terms and canceling out any common factors:

fβ€²(x)=lim⁑hβ†’02(x2+2hx+h2)βˆ’3(x+h)βˆ’2x2+3xhf'(x) = \lim_{h \to 0} \frac{2(x^2 + 2hx + h^2) - 3(x + h) - 2x^2 + 3x}{h}

fβ€²(x)=lim⁑hβ†’02x2+4hx+2h2βˆ’3xβˆ’3hβˆ’2x2+3xhf'(x) = \lim_{h \to 0} \frac{2x^2 + 4hx + 2h^2 - 3x - 3h - 2x^2 + 3x}{h}

fβ€²(x)=lim⁑hβ†’04hx+2h2βˆ’3hhf'(x) = \lim_{h \to 0} \frac{4hx + 2h^2 - 3h}{h}

Step 4: Evaluate the limit

Evaluate the limit as h approaches zero:

fβ€²(x)=lim⁑hβ†’04hx+2h2βˆ’3hhf'(x) = \lim_{h \to 0} \frac{4hx + 2h^2 - 3h}{h}

fβ€²(x)=lim⁑hβ†’0(4x+2hβˆ’3)f'(x) = \lim_{h \to 0} (4x + 2h - 3)

fβ€²(x)=4xβˆ’3f'(x) = 4x - 3

Therefore, the derivative of the quadratic function f(x) = 2x^2 - 3x is f'(x) = 4x - 3.

Example 2: Differentiating a Polynomial Function

Let's consider the polynomial function:

y=4x5βˆ’6x4+3xy = 4x^5 - 6x^4 + 3x

We want to find the derivative of this function using first principles.

Step 1: Write the definition of a derivative

The definition of a derivative is given by:

fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

Step 2: Substitute the function

Substitute the given function into the definition of a derivative:

fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

fβ€²(x)=lim⁑hβ†’04(x+h)5βˆ’6(x+h)4+3(x+h)βˆ’(4x5βˆ’6x4+3x)hf'(x) = \lim_{h \to 0} \frac{4(x + h)^5 - 6(x + h)^4 + 3(x + h) - (4x^5 - 6x^4 + 3x)}{h}

Step 3: Simplify the expression

Simplify the expression by combining like terms and canceling out any common factors:

fβ€²(x)=lim⁑hβ†’04(x5+5x4h+10x3h2+10x2h3+5xh4+h5)βˆ’6(x4+4x3h+6x2h2+4xh3+h4)+3(x+h)βˆ’4x5+6x4βˆ’3xhf'(x) = \lim_{h \to 0} \frac{4(x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5) - 6(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4) + 3(x + h) - 4x^5 + 6x^4 - 3x}{h}

fβ€²(x)=lim⁑hβ†’04x5+20x4h+40x3h2+40x2h3+20xh4+4h5βˆ’6x4βˆ’24x3hβˆ’36x2h2βˆ’24xh3βˆ’6h4+3x+3hβˆ’4x5+6x4βˆ’3xhf'(x) = \lim_{h \to 0} \frac{4x^5 + 20x^4h + 40x^3h^2 + 40x^2h^3 + 20xh^4 + 4h^5 - 6x^4 - 24x^3h - 36x^2h^2 - 24xh^3 - 6h^4 + 3x + 3h - 4x^5 + 6x^4 - 3x}{h}

fβ€²(x)=lim⁑hβ†’020x4h+40x3h2+40x2h3+20xh4+4h5βˆ’24x3hβˆ’36x2h2βˆ’24xh3βˆ’6h4+3hhf'(x) = \lim_{h \to 0} \frac{20x^4h + 40x^3h^2 + 40x^2h^3 + 20xh^4 + 4h^5 - 24x^3h - 36x^2h^2 - 24xh^3 - 6h^4 + 3h}{h}

Step 4: Evaluate the limit

Evaluate the limit as h approaches zero:

fβ€²(x)=lim⁑hβ†’020x4h+40x3h2+40x2h3+20xh4+4h5βˆ’24x3hβˆ’36x2h2βˆ’24xh3βˆ’6h4+3hhf'(x) = \lim_{h \to 0} \frac{20x^4h + 40x^3h^2 + 40x^2h^3 + 20xh^4 + 4h^5 - 24x^3h - 36x^2h^2 - 24xh^3 - 6h^4 + 3h}{h}

fβ€²(x)=lim⁑hβ†’0(20x4+40x3h+40x2h2+20xh3+4h4βˆ’24x3βˆ’36x2hβˆ’24xh2βˆ’6h3+3)f'(x) = \lim_{h \to 0} (20x^4 + 40x^3h + 40x^2h^2 + 20xh^3 + 4h^4 - 24x^3 - 36x^2h - 24xh^2 - 6h^3 + 3)

fβ€²(x)=20x4βˆ’24x3+3f'(x) = 20x^4 - 24x^3 + 3

Therefore, the derivative of the polynomial function y = 4x^5 - 6x^4 + 3x is y' = 20x^4 - 24x^3 + 3.

Example 3: Differentiating an Exponential Function

Let's consider the exponential function:

y=βˆ’x32+3y = -\frac{\sqrt[3]{x}}{2} + 3

We want to find the derivative of this function using first principles.

Step 1: Write the definition of a derivative

The definition of a derivative is given by:

fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

Step 2: Substitute the function

Substitute the given function into the definition of a derivative:

fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

fβ€²(x)=lim⁑hβ†’0βˆ’x+h32+3βˆ’(βˆ’x32+3)hf'(x) = \lim_{h \to 0} \frac{-\frac{\sqrt[3]{x + h}}{2} + 3 - \left(-\frac{\sqrt[3]{x}}{2} + 3\right)}{h}

Step 3: Simplify the expression

Simplify the expression by combining like terms and canceling out any common factors:

fβ€²(x)=lim⁑hβ†’0βˆ’x+h32+x32hf'(x) = \lim_{h \to 0} \frac{-\frac{\sqrt[3]{x + h}}{2} + \frac{\sqrt[3]{x}}{2}}{h}

f'(x) = \lim_{h \to 0} \frac{-\frac{\sqrt[3]{x + h} - \sqrt[3]{x}}{2}}{h}$<br/> **Differentiation Using First Principles: A Comprehensive Guide** ===========================================================

Q&A: Differentiation Using First Principles

Q: What is differentiation using first principles?

A: Differentiation using first principles is a method of finding the derivative of a function by using the definition of a derivative. The definition of a derivative is given by:

f&#x27;(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} </span></p> <h2><strong>Q: How do I apply the definition of a derivative to find the derivative of a function?</strong></h2> <p>A: To apply the definition of a derivative, follow these steps:</p> <ol> <li> <p><strong>Write the definition of a derivative</strong>: Write the definition of a derivative, which is given by:</p> <p class='katex-block'><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mi>f</mi><mo mathvariant="normal" lspace="0em" rspace="0em">β€²</mo></msup><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><munder><mrow><mi>lim</mi><mo>⁑</mo></mrow><mrow><mi>h</mi><mo>β†’</mo><mn>0</mn></mrow></munder><mfrac><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mi>h</mi><mo stretchy="false">)</mo><mo>βˆ’</mo><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow><mi>h</mi></mfrac></mrow><annotation encoding="application/x-tex">f&#x27;(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} </annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0519em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right:0.10764em;">f</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8019em;"><span style="top:-3.113em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">β€²</span></span></span></span></span></span></span></span></span><span class="mopen">(</span><span class="mord mathnormal">x</span><span class="mclose">)</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:2.1791em;vertical-align:-0.7521em;"></span><span class="mop op-limits"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.6944em;"><span style="top:-2.3479em;margin-left:0em;"><span class="pstrut" style="height:3em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mathnormal mtight">h</span><span class="mrel mtight">β†’</span><span class="mord mtight">0</span></span></span></span><span style="top:-3em;"><span class="pstrut" style="height:3em;"></span><span><span class="mop">lim</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.7521em;"><span></span></span></span></span></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.427em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord mathnormal">h</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord mathnormal" style="margin-right:0.10764em;">f</span><span class="mopen">(</span><span class="mord mathnormal">x</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord mathnormal">h</span><span class="mclose">)</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">βˆ’</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord mathnormal" style="margin-right:0.10764em;">f</span><span class="mopen">(</span><span class="mord mathnormal">x</span><span class="mclose">)</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.686em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span></span></span></span></span></p> </li> <li> <p><strong>Substitute the function</strong>: Substitute the given function into the definition of a derivative.</p> </li> <li> <p><strong>Simplify the expression</strong>: Simplify the expression by combining like terms and canceling out any common factors.</p> </li> <li> <p><strong>Evaluate the limit</strong>: Evaluate the limit as h approaches zero.</p> </li> </ol> <h2><strong>Q: What are some common mistakes to avoid when differentiating using first principles?</strong></h2> <p>A: Some common mistakes to avoid when differentiating using first principles include:</p> <ul> <li><strong>Not simplifying the expression</strong>: Failing to simplify the expression can lead to incorrect results.</li> <li><strong>Not evaluating the limit</strong>: Failing to evaluate the limit can lead to incorrect results.</li> <li><strong>Not using the correct definition of a derivative</strong>: Using the wrong definition of a derivative can lead to incorrect results.</li> </ul> <h2><strong>Q: Can I use differentiation using first principles to find the derivative of a complex function?</strong></h2> <p>A: Yes, you can use differentiation using first principles to find the derivative of a complex function. However, it may be more challenging to simplify the expression and evaluate the limit.</p> <h2><strong>Q: Are there any other methods of finding the derivative of a function?</strong></h2> <p>A: Yes, there are other methods of finding the derivative of a function, including:</p> <ul> <li><strong>Power rule</strong>: The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1).</li> <li><strong>Product rule</strong>: The product rule states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x).</li> <li><strong>Quotient rule</strong>: The quotient rule states that if f(x) = u(x)/v(x), then f'(x) = (u'(x)v(x) - u(x)v'(x)) / v(x)^2.</li> </ul> <h2><strong>Q: When should I use differentiation using first principles?</strong></h2> <p>A: You should use differentiation using first principles when:</p> <ul> <li><strong>The function is not in a form that can be easily differentiated using other methods</strong>: If the function is not in a form that can be easily differentiated using other methods, such as the power rule or product rule, then you may need to use differentiation using first principles.</li> <li><strong>You need to find the derivative of a complex function</strong>: If you need to find the derivative of a complex function, then you may need to use differentiation using first principles.</li> </ul> <h2><strong>Conclusion</strong></h2> <p>Differentiation using first principles is a powerful tool for finding the derivative of a function. By following the steps outlined in this article, you can use differentiation using first principles to find the derivative of a wide range of functions. Remember to simplify the expression and evaluate the limit carefully to ensure accurate results.</p>